circuit analysis

what would be the value of Vc(zero minus/plus) ie Vc(0-)=Vc(0+) . Is it [3*48/(48+100) =]97.30 or [3*48=]144 . I cant understand why the answer 97.30 is given in the book Hayt and Kimmerly 8th edison-- the source is a current source- not a voltage source, how the voltage division formula is applied here.

If for above 97.30 is true then how come for the circuit shown below answer for Vc(0-)=48:
According to the author Vc(0-) needs to be [3*48/(48+24)=] 2 rather than 48 [ =3*{24/(24+48)}*48] right?

which will result in an answer that has units of current, not units of voltage.

To find the value of Vc(0-), analyze the circuit as it exists prior to t=0. Since the current source will have been at 3A for a very long time and nothing else is changing, what will the capacitor and inductor behave like at that time? So what does the effective circuit look like?

In addition, you need to always ask if the answer makes sense. Does it seem reasonable that that expression can yield a value that is almost 100? You basically have

Thanks for the advice WBahn; we need to keep an eye on the "units". Just troubling you a bit more, can u plz let me know whether the 3 calculations [shown below] are right or wrong for the given circuit
So (current through resistor) --> Ir(0-) = 3*48/(48+100) = 0.973 A
and Vc(0-) = 3*48= 144 V
and (current through inductor) -->Il(0-) = 3*100/(48+100) = 2.027 A

And......plz have a look at this note:
At t=0- ; the INDUCTOR acts as short circuit and CAPACITOR as open circuit to DC
At t=0+ ; INDUCTOR replaced with current source[ Il(0+)= Il(0-) ] and CAPACITOR replaced with voltage source[ Vc(0+) = Vc(0-) ] and have to remove 48ohm and also the 3A current source.....WHY we need to remove this 3A curr source at t=0+ ?